Garciano, Agnes D.

Hiramine, Yutaka

Yokonuma, Takeo

Kyoto University Research Information Repository

TitleRelative Difference Sets in Dihedral Groups (AlgebraicCombinatorics)Author(s)Garciano, Agnes D.; Hiramine, Yutaka; Yokonuma, TakeoCitation数理解析研究所講究録 (2005), 1440: 74-81Issue Date2005-07URL http://hdl.handle.net/2433/47531RightType Departmental Bulletin PaperTextversionpublisherKyoto University74Relative Difference Sets in Dihedral GroupsAgnes D. GarcianoMathematics Department, Ateneo de Manila UniversityLoyola Heights, Quezon City, PhilippinesYutaka HiramineDepartment of Mathematics, Faculty of Education, Kumamoto UniversityKurokami, Kumamoto, JapanTakeo YokonumaDepartment of Mathematics, Sophia UniversityKioichi, Chiyoda-ku, Tokyo, Japan1. IntroductionA $(m, n, k, \lambda)$ relative difference set (RDS) in a finite group $G$ of order mnrelative to a subgroup $N$ of order $n$ , is a $k$ -element subset $R$ of $G$ wherein everyelement of $G-N$ has exactly A representations as $r_{1}r_{2}^{-1}$ with $r_{1}$ , $r_{2}\in R$ . Moreover,no nonidentity element of $N$ has such a representation. $N$ is called the forbiddensubgroup. If for a subset $X$ of $G$ , we identify $X$ with the group ring element $X=$$\sum_{x\epsilon X}x$$\in$ $\mathrm{C}[\mathrm{G}]$ and set $X^{(-1)}= \sum_{x\epsilon X}x_{1}^{-1}$ then $R$ is a $(m, n, k, \lambda)$ RDS in $G$ relativeto $N$ if $RR^{\langle-1)}=k+\lambda(G-N)$ . It follows that $k(k-1)$ $=\lambda n(m-1)$ . Note that if$N=1$ , then $R$ is an $(m, k, \lambda)$ difference set in the usual sense.The notion of a relative difference set was introduced by Elliot and Butson [1].The following result which is due to them is fundamental in the study of RDS’s.Result 1.1. ([1]) Let $R$ be a $(m, n, k, \lambda)$ relative difference set in a group $G$ relative toa subgroup $N$ and let $U$ be a normal subgroup of $G$ contained in $N$ . If $\phi$ : $Garrow G/U$is the canonical epimorphism and $|U|=u$ , then $\phi(R)$ is a $(m,$ $\frac{n}{u}$ , $k$ , $u\lambda)$ relativedifference set in $\overline{G}(=G/U)$ with respect to $\overline{N}(=N/U)$ .In particular, if $N=U$, then $\phi(R)$ is a $(m, k, n\lambda)$ ordinary difference set in$\overline{G}(=G/N)$ . We may then consider $R$ as an ” extension” of an ordinary difference set.Although trivial ordinary difference sets with parameters of the form $(v+2,$ $v1$$1$ , $v)$ and $(v, v, v)\backslash$ , $v>0$ , exist in any group, it is still a question whether or notextensions of these difference sets also exist. In dihedral groups for instance, it isconjectured that only trivial ordinary difference sets exist. Hence, a problem thatwe would like to consider is whether extensions of these trivial difference sets existin dihedral groups.A relative difference set in a group $G$ is said to be semiregular or of affine typeif its parameters are of the form $(\mathrm{n}\mathrm{A}3n, n\lambda, \lambda)$ or $(n\lambda +2, n, n\lambda +1, \lambda)$ respectively1440 2005 74-8175If $N$ is a normal subgroup of $G$ and $R$ is either a semiregular or affine type RDSin $G$ , then $\overline{R}$ is a trivial ordinary difference set in $\overline{G}$ by Result 1.1. We say that a$(m, n, k, \lambda)$ relative difference set is trivial if $k=1$ or $(n, k)\in\{(1, m), (1, m-1)\}$ .If the conjecture mentioned above is true, then the only nontrivial RDS’s thatcan exist in dihedral groups relative to a normal subgroup are either semiregular orof affine typeThe only nontrivial relative difference set up to equivalence in a dihedral groupknown to the authors is as follows:Example 1.2. Let $G=\langle x, y|x^{4}=y^{2}=1, y^{-1}xy=x^{-1}\rangle$ be the dihedral group oforder 8. Then $D=\{1, xy, x^{2}y, x^{3}\}$ is a (4, 2, 4, 2) relative difference set in $G$ relativeto $\langle y\rangle$ .In [2], the following was shown.Result 1.3. ([2]) There exists no nontrivial semiregular relative difference set in anydihedral group relative to a normal subgroup.In section 3, we prove the following.Theorem 3.1. There is no relative difference set of affine type in dihedral groups.2. PreliminariesWe will use the following results which we mention here without proof.Result 2.1. ([3]) Let $X$ be an $n\mathrm{x}n$ circulant matrix$X=\ovalbox{\tt\small REJECT}^{x_{n-1}}x_{0}x_{1}..\cdot$ $x_{1}x_{0}x_{2}..\cdot$ .$\cdot$.$x_{n-2}x_{n-1}x_{0}..\cdot\ovalbox{\tt\small REJECT}$Then $\det(X)$ $= \prod_{0\leq i\leq n-1}(x_{0}+\xi^{i}x_{1}+\xi^{2i}x_{2}+\cdots+\xi^{(n-1)i}x_{n-1})$ , where 4is aprimitive$n\mathrm{t}\mathrm{h}$ root of unity. Moreover, if $\det(x)$ $\neq 0$ , then $X^{-1}$ is also circulant.Result 2.2. ([4]) (Inversion Formula). Let $G$ be an abelian group and $A= \sum_{g\in G}\alpha_{g}g$be an element of the group $\mathrm{a}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a}\mathbb{C}[\mathrm{G}]$ . Then, $\alpha_{g}=\frac{1}{|G|}\sum_{\chi\in G^{*}}\chi(A)\chi(g^{-1})$ for each$g$ $\in G$ where $G^{*}$ is the group of characters of $G$ .Throughout the rest of this paper, we will assume the following76Assumptions. Let $R$ be a $(n\lambda+2,n,n\lambda+1, \lambda)$ (A $>0$ ) relative difference set in adihedral group $G$ relative to a subgroup $N$ . Set $G=C\langle t\rangle$ where $C$ is a cyclic groupand $t$ is an involution which inverts $C$ . Set $R=A+Bt$ where $A$ and $B$ are subsetsof $C$ . By exchanging $Rt$ for $R$ if necessary, we may assume $|A|\leq|B|$ .Proposition 2.3. Under the above assumptions, the following hold:(i) If $N\subset C$ , then $AA^{(-1)}+BB^{(-1)}=(n\lambda+1)+\lambda(C-N)$ and $AB= \frac{\lambda}{2}C$ .Furthermore, $|A|= \frac{n\lambda}{2}$ and $|B|= \frac{n\lambda}{2}+1$ .(ii) If $N\not\subset C$ , $N_{1}=N\cap C$ and $N_{2}=Nt\cap C$ , then $AA^{(-1)}+BB^{(-1)}=(n\lambda+$$1)+\lambda(C-N_{1})$ and $AB= \frac{\lambda}{2}$ (C-N2). Furthermore, $|A|= \frac{(n\lambda+1)-\sqrt{n\lambda+1}}{2}$and $|B|= \frac{(n\lambda+1)+\sqrt{n\lambda+1}}{2}$ .Proof. We have $RR^{(-1)}=(A+Bt)(A^{(-1)}+tB^{(-1)})=AA^{(-1)}+BB^{(-1)}+2ABL$Suppose $N\subset \mathrm{G}$ . By definition, $RR^{(-1)}=(n\lambda+1)+\lambda(C+Ct-N)$ .Thus, $AA^{(-1)}+BB^{(-1)}=(n\lambda+1)+\lambda(C-N)$ and $AB= \frac{\lambda}{2}$ $C$ . If $|A|=a$and $|B|=b$ , it follows that $a+b=n\lambda+1$ and $ab= \frac{\lambda}{4}n(n\lambda+2)$ . Hence (i)holds.Suppose $N\not\subset C$ . Then, $RR^{(-1)}=(n\lambda+1)+\lambda(C+Ct-N_{1}-N_{2}t)$ .Thus, $AA^{(-1)}+BB^{(-1)}=(n\lambda+1)+\lambda(C-N_{1})$ and $AB= \frac{\lambda}{2}(C-N_{2})\lambda$ . If$|A|=a$ and $|B|=b$ , it follows that $a+b=n\lambda+1$ and $ab=\overline{4}n(n\lambda+2)$ .Hence (ii) holds. $\bullet$3. Nonexistence of Affine Type Relative Difference Sets in DihedralGroupsTo prove our main theorem, we first show a necessary condition on the forbiddensubgroup.Proposition 3.1. Let R be a relative difference set of afHne type in a dihedral groupG relative to a subgroup N of G. Then, N is normal in G.We will prove Proposition 3,1 in Lemmas 3.2 - 3.7. As mentioned in the previoussection, we let $G=C\langle t\rangle$ where $C$ is a cyclic subgroup of $G$ and $t$ is an element of $G$which inverts $C$ . Set $R=A+Bt$ where $A$ and $B$ are subsets of $C$ .77Suppose the proposition is false and let $G$ be a minimal counterexample to theproposition. As every element outside $C$ is an involution and inverts $C$ , we mayassume that $t\in N$ .Lemma 3,2. n $=2$ . In particular,(i) $G=CN$ , $N=\langle t\rangle$ and $C\cong Z_{2(\lambda+1)}$(ii) $D$ is a $(2\mathrm{A}+2,2,2\lambda+1, \lambda)$ relative difference in $G$ with respect to $N$ .Proof. Let $L=N\cap C$ . Then $[N:L]=2$ and $G\triangleright L$ as $C$ is cyclic. Therefore, byResult 1.1, $\overline{D}$ is a difference set with parameters $(n\lambda+2,2,$ $n\lambda+1$ , $\frac{n\lambda}{2})$in $\overline{G}(=G/L)$ relative to $\overline{N}(=N/L\cong \mathbb{Z})$ . Clearly, $\overline{G}\beta\overline{N}$. By the minimality of $G$ , $L=1$ . Thus $N=\langle t\rangle$ . @By Proposition 2.3, we have$AA^{(-1)}+BB^{(-1)}=(2\lambda+1)+\lambda(C-1)$ (1)$AB= \frac{\lambda}{2}(C-1)$ (2)$|A|= \frac{2\lambda+1-\sqrt{2\lambda+1}}{2}$ and $|B|= \frac{2\lambda+1+\sqrt{2\lambda+1}}{2}$ (3)Lemma 3.3. We may assume that C $=A+B^{(-1)}+1$ .Proof. By (2) and (3), $A\cap B^{(-1)}=\phi$ and $|A|+|B|=|C|-1$ . Hence $C=A\mathrm{U}B^{(-1)}$$\cup\{g\}$ for some $g\in C$ . Exchanging $D$ for $Dg^{-1}$ if necessary, we may assumethat $g=1$ .$\bullet$Lemma 3.4. A $=A^{(-1)}$ and B $=B^{(-1)}$ .Proof. Let $\chi$ be a nonprincipai character of $C$ and set $\chi(A)=a$ and $\chi(B)=b$ . By(1), (2) and Lemma 3.3, $a\overline{a}+b\overline{b}=\lambda+1$ , $ab= \frac{-\lambda}{2}$ and $a+\overline{b}+1=0$ . Itfollows that $2\mathrm{a}\mathrm{a}+a+\overline{a}=$ A $=2\mathrm{a}\mathrm{a}+2a$ . Hence $a=\mathrm{a}$ . By Result 2.2,$A-A^{(-1)}=0$ . Thus, $A=A^{(-1)}$ and as $B=C$ - $A-1$ , we have $B=B^{(-1)}$ .$\bullet$By (3) above, we can set $2s+1=\sqrt{2\lambda+1}$ for a positive integer $s$ .Then $\lambda=2s^{2}+2s$ and $D$ is a $(4s^{2}+4s+2,2, (2s+1)^{2},2s^{2}+2)$ RDSin $G(\cong D_{4(2s^{2}+2s+1)})$ . Moreover, by (1), (2) and Lemma 3.4, we have$|A|=2s^{2}+s$ , $|B|=2s^{2}+3s+1$ , $C=A+B+1$ , $A^{2}+B^{2}=$$(2s +1)^{2}+(2s^{2}+2)(C-1)$ , $AB=(s^{2}+s)(C-1)$ . Hence, the followinghold78Lemma 3.5. $A^{2}+A=s^{2}C+s^{2}+s$ , $B^{2}+B=(s+1)^{2}C+s^{2}+s$ .Let $M$ be the unique subgroup of $C$ of index 2 and let $d$ be an involution of $C$ .Then $C=M\mathrm{x}$ $\langle d\rangle$ . Set $\overline{C}=C/M(=\{\overline{1}, \overline{d}\})$ .Lemma 3.6. $|A\cap M|=s^{2}+s$ , $|A\cap Md|=s^{2}$ , $|B\cap M|=s^{2}+s$ , $|B\cap Md|$ $=(s+1)^{2}$ .Proof. Let $v=|A\cap M|$ and $w=|A\cap Md|$ . Then $\overline{A}=v+w\overline{d}$ and $v+w=|A|=2s^{2}+s$ .By Lemma 3.5, $(v+w\overline{d})^{2}+(v+w\overline{d}\rangle$ $=s^{2}(2s^{2}+2s+1)(\overline{1}+\overline{d})+s^{2}+s$ . Itfollows that $v^{2}+w^{2}+v=2s^{4}+2s^{3}+2s^{2}+s$ and $2vw+w=2s^{4}+2s^{3}+s^{2}$and so $(v-w)^{2}+(v-w)$ $=s^{2}+s$ . Thus, $v-w$ $=s$ or $v-w=-(s+1)$ . If$v-w=-(s+1)$ , then $2w=2s^{2}+2s+1$ , a contradiction. Hence $v-w=s$and so $v=s^{2}+s$ , $w=s^{2}$ . The other equations in the Lemma can beproven similarly. ElLemma 3.7. s is even.Proof. By Lemma 3.6, $B\cap Md\neq\phi$ . Let $g\in B\cap Md$ and set $\Omega=\{(x, y)|x$ , $y\in B$ ,$g=xy\}$ . By Lemma 3.5, $|\Omega|=(s+1)^{2}-1$ . If $s$ is odd; then $|\Omega|\equiv 1$(mod 2). As $(x, y)\in\Omega$ implies $(y, x)\in\Omega$ , there is an element $z\in B$ suchthat $(z, z)\in\Omega$ . Thus, $g=z^{2}\in Md$ , a contradiction. Thu $\mathrm{s}$ , $s$ is even. $\bullet$Proof of Proposition 3.1:By Lemma 3.7, $s=2\ell$ for some integer $\ell>0$ . By Lemma 3.6, $A\cap Md\neq\phi$ . Let$g\in BnMd$ and set $\Omega=\{(x, y)|x, y\in A, xy=g\}$ . By Lemma 3.5, $|\Omega|=s^{2}-1\equiv 1$(mod 2). By a similar argument as in Lemma 3.7, we have a contradiction. Thus,$G\triangleright N$ .$\bullet$Proposition 3.8. Let G be a dihedral group and N a normal subgroup of G. Then,there is no nontrivial relative difference set of affine type in G relative to N.In the rest of this section, let $G$ be a minimal counterexample to Proposition3.8 and let $R$ be a $(\mathrm{p}\mathrm{X}+2, p, p\lambda+1, \lambda)$ RDS in C. By the minimality condition,$p$ is a prime. As mentioned in Section 2, we let $C-=\mathrm{C}\{\mathrm{t}$) where $t$ inverts the cyclicgroup $C$ and let $R=A+Bt$ where $A$ and $B$ are subsets of $C$ . Exchanging $R$ for itstranslate, if necessary, we may assume $R\cap N=\phi$ and $R\cup\{1\}$ is a complete set ofcoset representatives of $G/N$ . Since $G\triangleright N$, $N$ is contained in $C$ . By Proposition 2.3,we have$AA^{(-1)}+BB^{(-1)}=(p\lambda+1)+\lambda(C-N)$ (4)$AB= \frac{\lambda}{2}C$ (5)$|A|= \frac{\lambda}{2}p$ , $|B|= \frac{\lambda}{2}p+1$ (6)$7\theta$Let $h= \frac{\lambda}{2}p+1$ . Moreover, let $C=HN$ where $N=\langle s\rangle$ $\cong \mathbb{Z}_{\mathrm{p}}$ and $H\cong \mathbb{Z}_{\mathrm{h}}$ .Thus, we can set$A=A_{0}+A_{1}s+\cdots+A_{p-1}s^{p-1}$ , $B=B_{0}+B_{1}s+\cdots+B_{p-1}s^{p-1}$ (7)for some subsets $A_{0}$ , . . . ’ $A_{p-1}$ , $B_{0}$ , $\ldots$ , $B_{p-1}$ of $H$ .Lemma 3.9. The following hold:(i) $A_{i}\cap A_{j}=B_{i}\cap B_{j}=\phi$ $\forall \mathrm{i}$ , $j$ with $0\leq \mathrm{i}$ , $j\leq p-1$ , $\mathrm{i}\neq i$ .(ii) $H=1+ \sum_{\mathit{0}\leq i\leq p-1}A_{i}=\sum_{0\leq i\leq p-1}B_{i}$ .Proof. Since $N=\langle s$ } and $AA^{(-1)}\cap N=BB^{(-1)}\cap N=\{1\}$ by (4 $\grave{}, (\mathrm{i})$ holds. Hence,$|A|= \sum_{0<\mathrm{i}\leq p-1}|A_{i}|$ and $|B|= \sum_{0\leq i\leq \mathrm{p}-1}|B_{i}|$ . By (6), $|A|=h-1$ and$|B|=h$ . Then, (ii) follows immediately, $\bullet$Substituting (7) into equations (4) and (5), we have$A_{0}B_{0}+A_{1}B_{p-1}+A_{2}B_{p-2}+ \cdots+A_{p-1}B_{1}=\frac{\lambda}{2}H$$A_{0}B_{1}+A_{1}B_{0}+A_{2}B_{p-1}+ \cdots+A_{p-1}B_{2}=\frac{\lambda}{2}H$$A_{0}B_{l}[perp] A_{1}B_{i-1}+A_{2}B_{i-2}+ \cdots+A_{p-1}B_{i-p+1}=\frac{\lambda}{2}H$ (8). . . $+$ $\cdot$ . . $+$ $\cdot$ . . $+$ $\cdot$ . . $= \frac{\lambda}{2}H$$A_{0}B_{p-1}+A_{1}B_{p-2}+A_{2}B_{p-3}+ \cdots+A_{p-1}B_{0}=\frac{\lambda}{2}H$and$A_{0}A_{p-1}^{(-1)}+A_{1}A_{0}^{(-1)}+A_{2}A_{1}^{(-1)}+\cdots+A_{p-2}A_{p-3}^{(-1)}+A_{p-1}A_{p-2}^{(-1)}$$+B_{0}B_{p-1}^{(-1)}+B_{1}B_{0}^{(-1)}+B_{2}+B_{1}^{(-1)}+\cdots+B_{p-1}B_{p-2}^{(-1)}=\lambda(H-1)(9)$Let $\chi$ be a character of $H$ . By (8), we have$\ovalbox{\tt\small REJECT}_{\chi(B_{p-2})}^{\chi(B_{0})}\chi(B_{p-1}^{\cdot}.\cdot)\chi(B_{1})$$\chi(B_{p-1})\chi(B_{p-3})\chi(B_{p-2})\chi(B_{0})$ .$\cdot$.$\chi(B_{p-1})\chi(B_{2})\chi(B_{1})\chi(B_{0})\ovalbox{\tt\small REJECT}$ $\ovalbox{\tt\small REJECT}_{\chi(A_{p-2})}^{\chi(A_{0})}\chi(A_{p-1}^{\cdot}.\cdot)\ovalbox{\tt\small REJECT}\chi(A_{1})=\frac{\lambda}{2}\chi(H)\ovalbox{\tt\small REJECT}_{1}^{1}11^{\cdot}\cdot$ . $\ovalbox{\tt\small REJECT}$ (10)Lemma 3.10. The following hold.(i) $|A_{0}|=|A_{1}|= \cdots=|A_{p-1}|=\frac{\lambda}{2}$ .80(ii) $|B_{0}||B_{p-1}|+|B_{1}||B_{0}|+ \cdots+|B_{p-1}||B_{p-2}|=\frac{p\lambda^{2}}{4}$ .Proof. Let $|A_{i}|=a_{i}$ and $|B_{i}|=b_{i}$ for i $=0$ , 1, \ldots ,p-1. If $\chi$ is the principalcharacter of H, then by (10),$\ovalbox{\tt\small REJECT}_{b_{p-1}}^{b_{0}}b_{p-2}b_{1}..\cdot$ $b_{p-1}b_{p-3}b_{p-2}b_{0}$$b_{p-1}b_{2}b_{0}b_{1}\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}_{a_{p-1}}^{a_{0}}a_{p-2}..\cdot\ovalbox{\tt\small REJECT} a_{1}=\frac{\lambda h}{2}$$\ovalbox{\tt\small REJECT}_{1}^{1}11^{\cdot}.\cdot\ovalbox{\tt\small REJECT}$ (11)Let $P$ be the $p\mathrm{x}$ $p$ matrix in the above equation (11), By Result 2.1,$\det(F)=\Pi_{0\leq i\leq p-1}(b_{0}+b_{p-1}\zeta^{i}+b_{\mathrm{p}-2}\zeta^{2i}+\cdots+b_{1}\zeta^{(p-1)i})$ , where $\langle$ is aprimitive p-th root of unity. By Lemma 3.9 (ii), $b_{\mathit{0}}+b_{p-1}+b_{p-2}+\cdots+$$b_{1}=|H|\neq 0$ . Suppose $b_{0}+b_{p-1}\zeta^{\dot{\mathrm{t}}}+b_{p-2}\zeta^{2i}+\cdots+b_{1}\zeta^{(p-1)i}=0$ forsome $i\neq 0$ . Set $\theta=\zeta^{i}$ . Then, 0 is a primitive p-th root of unity and$x^{p-1}+x^{p-2}+\cdots+x+1$ is a minimal polynomial of 71 over (Q. Hence$b_{0}=b_{p-1}=b_{p-2}=\cdots=b_{1}$ . However, $pb_{0}= \sum_{0\leq i\leq p-1}b_{i}=|H|=\frac{\lambda}{2}p+1$ ,a contradiction. Thus $\det(P)$ $\neq 0$ . By Result 2.1, $P^{-1}$ is also circulant.Since $(a_{0}, a_{1}, \ldots, a_{p-1})^{T}=\frac{\lambda h}{2}P^{-1}(1,1, \ldots 1)^{T}$ , it follows that $a0=a_{1}=$$\ldots=a_{p-1}$ . Hence, by Lemma 3.9, $a_{0}=a_{1}= \cdots=a_{p-1}=\frac{\lambda}{2}$ . Thus (i)holds and (ii) follows from (9) and (i).Lemma 3.11. Let $\chi$ be a non-principal character of $H$ . Then $\chi(B_{0})=\chi(B_{1})=$. . . $=\chi(B_{p-1})=0$ .Proof. Set $\chi(A_{i})=\alpha_{i}$ and $\chi(B_{i})=\beta_{i}$ for $\mathrm{i}=0,1$ , $\ldots$ , $p-1$ . By (10)$\ovalbox{\tt\small REJECT}_{\alpha_{p-1}}^{\alpha_{0}}\alpha_{p-2}\alpha_{1}..\cdot$$\alpha_{p-2}\alpha_{p-3}\alpha_{p-1}\alpha_{0}$$\alpha_{p-1}\alpha_{0}\alpha_{2}\alpha_{1}\ovalbox{\tt\small REJECT}$ $\ovalbox{\tt\small REJECT}_{\beta_{p-1}}^{\beta_{0}}\beta_{p-2}..\cdot\ovalbox{\tt\small REJECT}\beta_{1}=\ovalbox{\tt\small REJECT}$$0^{\mapsto}$0.$\cdot$.00(12)Let $Q$ be the $p\mathrm{x}p$ matrix in the equation (12) above. By Result 2.1,$\det(Q)=\Pi_{0\leq i\leq p-1}(\alpha_{0}+\alpha_{p-1}^{\zeta^{i}}+\alpha_{p-2}\zeta^{2i}+\cdots +\alpha_{1}\zeta^{(p-1)i})$ , where $\langle$ is aprimitive pit root of unity. Let $\mathrm{i}\neq 0$ , $\mathrm{i}\in$ { $0,$ 1, $\ldots$ ,p– 1} and let $\eta=$$\alpha_{0}+\alpha_{p-1}\theta+\alpha_{p-2}\theta^{2}+\cdots+\alpha_{1}\theta^{p-1}$ where $\theta=\zeta^{\mathrm{i}}$ . Then, we have $\eta^{p}=\alpha_{0}^{p}+$$\alpha_{p-1}^{p}+\alpha_{p-2}^{p}+\cdots+\alpha_{1}^{p}$ (mod $p$) $\equiv\sum_{0\leq i\leq p-1}(\sum_{x\in A_{i}}\chi(x))^{p}$ (mod $p$) $\equiv$$\sum_{0\leq i\leq p-1}(\sum_{x\in A_{i}}\chi(x^{p}))=\sum_{x\in\{A_{0},..,A_{p-1}\}}\chi(X^{(p)})=\chi(H^{(p)}-1)$ byLemma 3.9 (ii). On the other hand, $H^{(p)}=H$ as $(p, h)=1$ . Hence $\eta^{p}\equiv-1$81(mod $p$) and so $\eta^{p}=-1+p\alpha$ for an algebraic integer $\alpha\in \mathbb{Z}[\theta]$ . If $\eta=0$ ,then $\alpha=\frac{1}{p}$ , a contradiction. Hence $\det(Q)\neq 0$ . Thus, the lemma holds.Proof of Proposition 3.8:By Lemma 3.11 and Result 2,2, there exist $c_{0}$ , $c_{1}$ , $\ldots$ , $c_{p-1}\in oe$ such that $B_{0}=$$\mathrm{C}\mathrm{q}\mathrm{H}$ , $B_{1}=c_{1}H$, . .. , $B_{p-1}=c_{p-1}H$ . Since each $B_{i}$ is a subset of $H$ , $B_{i_{0}}=H$ and$B_{l}=\phi(\forall i\neq \mathrm{i}_{0})$ for some $\mathrm{i}_{0}\in\{0,1, \ldots,p-1\}$ . By Lemma 3.10, $p \frac{\lambda^{2}}{4}=0$ . Thus$\lambda=0$ , a contradiction. $\bullet$By Propositions 3.1 and 3.8, we have the following.Theorem 3.12. There is no nontrivial relative difference set of affine type in dihedralgroups.References[1] J.E. Elliot and A.T. Butson, Relative difference sets, Illinois J. Math, vol10(1966), pp. 517 - 531.[2] D.T. Elvira and Y. Hiramine, On Non-Abelian Semi-Regular Relative DifferenceSets, Finite Fields and Applications, Springer, Berlin, 2001, pp. 122 - 127.[3] W. Greub, Linear Algebra, fourth edition, Springer, (1975).[4] A. Pott, Finite Geometry and Character Theory, Lecture Notes in Mathematics1601, Springer-Verlag, Berlin (1995)

Relative Difference Sets in Dihedral Groups (Algebraic Combinatorics)

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Relative Difference Sets in Dihedral Groups (Algebraic Combinatorics)

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